Logic Questions and Answers

Answer: A

Correct answer: JAK

The first letters advance by one through the series: F, G, H, I, so the next is J. The middle letter is A throughout. The last letters run G, F, I, H, so they advance by two and alternate high and low within each pair; the pair beginning with J takes K. The answer is therefore JAK.

Answer: C

Correct answer: 135 chalk sticks

Dividing by half is dividing by one half, which doubles the quantity: \[ \frac{60}{1/2} = 120, \qquad 120 + 15 = 135. \] The trap is to read it as dividing in half, which would give 30 and then 45. The wording is the whole of the question.

Answer: D

Correct answer: Both (a) & (b)

The first is the absorption law: \( (P \lor Q) \land Q \) simplifies to Q, so the biconditional is \( Q \leftrightarrow Q \) and is always true. The second is the disjunctive syllogism: given \( P \lor Q \) and \( \lnot P \), Q must hold. The third fails when P is true and Q false, since the antecedent is then true and the consequent false.

Answer: C

Correct answer: Logically equivalent to p ^ q

Distribute: \( p \land (\lnot p \lor q) \equiv (p \land \lnot p) \lor (p \land q) \), and the first disjunct is a contradiction, so the whole reduces to \( p \land q \). It is therefore neither a tautology nor a contradiction - it is true exactly when both p and q are true.

Answer: B

Correct answer: Father-in-law

Work inward from the description. Smith’s sister’s mother is Smith’s own mother. Her grandson, on Smith’s side of the family, is Smith’s son. Lilly is that son’s wife, so Smith is her husband’s father - her father-in-law.

Answer: A

Correct answer: (P ^ Q) → R

Expand both sides into disjunctive form. The left is \( \lnot P \lor (\lnot Q \lor R) \), which is \( \lnot P \lor \lnot Q \lor R \); the right is \( \lnot(P \land Q) \lor R \), which by De Morgan is the same expression. The result is the exportation law, and it is why a nested implication can be flattened into one with a conjunctive antecedent.

Answer: D

Correct answer: Q ^ (P v R)

Work forward from the premises. From \( P \to M \) and \( \lnot M \), modus tollens gives \( \lnot P \). With \( P \lor Q \) that gives Q, and \( Q \to R \) then gives R. So Q and R are both true, and since R is true so is \( P \lor R \) - which makes the conjunction the valid conclusion.

Answer: B

Implication associates to the right, so the formula reads

\[ (A \wedge B) \rightarrow \big((\neg C \wedge A) \rightarrow (A \equiv 1)\big) \]

A formula is valid (a tautology) when it is true under every assignment. Take any assignment and consider the outer implication:

• If \( A \wedge B \) is false, the outer implication is true by vacuous truth - an implication with a false antecedent is always true.
• If \( A \wedge B \) is true, then \( A \) is true, so \( A \equiv 1 \) is true. The inner implication therefore has a true consequent and is true regardless of \( \neg C \wedge A \), which makes the outer implication true as well.

Both cases give true, so the formula holds under all \( 2^{3} = 8 \) assignments of \( A, B, C \). It is valid.

That also rules out the other options: a contradiction is false everywhere, and "satisfiable but not valid" requires at least one assignment making it false. \( C \) never affects the outcome - it appears only inside a subformula whose value is irrelevant once \( A \) is known.

Answer: B

Correct answer: ~p v q

This is the definition of material implication: \( p \to q \) is false only when p is true and q false, which is exactly when \( \lnot p \lor q \) is false. The equivalence is what allows implications to be eliminated from a formula so that De Morgan’s laws and distribution can be applied.

Answer: D

Correct answer: KLLA

Take the four positions separately. The first letters go E, G, I, ?, M, advancing by two, so the missing one is K. The second is L throughout and the fourth is A throughout. The third letters go F, H, J, ?, N, also advancing by two, so the missing one is L. That gives KLLA.